Phenol Electrophilic Substitution

8 MCQs9-step worked example
Source: NCERT BiomoleculesPYQ coverage: NEET 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026

Lesson

The –OH group on a benzene ring activates the ring powerfully toward electrophilic substitution. This activation is the core of phenol reactivity in NEET, and the common confusion — treating –OH as just another activating group without accounting for the resonance-versus-induction balance in substituted phenols — costs marks reliably.

Why phenol is highly activated. The oxygen lone pair delocalises into the ring, increasing electron density at ortho and para positions. This makes phenol far more reactive than benzene toward electrophiles. NCERT Class 12 Chemistry Chapter 7 (Alcohols, Phenols and Ethers), page 14, documents this activation and the resulting ortho/para directing nature.

Bromination — the classic test. Phenol reacts with bromine water (Br₂/H₂O) at room temperature to give 2,4,6-tribromophenol (a white precipitate). No Lewis acid catalyst is needed — contrast this with benzene, which requires FeBr₃. This difference is a direct NEET question target. For mono-bromination, use Br₂ in CS₂ at low temperature, which gives predominantly para-bromophenol due to steric preference.

Nitration — temperature controls the product. Dilute HNO₃ at low temperature gives a mixture of ortho- and para-nitrophenol. Concentrated HNO₃ or a mixture of conc. HNO₃ + conc. H₂SO₄ gives 2,4,6-trinitrophenol (picric acid).

Kolbe's reaction (electrophilic carboxylation). Sodium phenoxide treated with CO₂ under pressure at 125°C gives sodium salicylate (ortho-hydroxybenzoic acid) after acidification. This is an electrophilic substitution unique to phenols.

Reimer-Tiemann reaction. Phenol + CHCl₃ + NaOH gives salicylaldehyde (ortho-hydroxybenzaldehyde) via an intermediate dichlorocarbene electrophile.

The substituted-phenol trap. When NEET asks you to predict the product of electrophilic substitution on a substituted phenol, you must consider both the –OH directing effect and the substituent's effect. Ignoring the resonance contribution and relying solely on inductive effects leads to wrong position predictions — this is a documented distractor pattern (observed in 2021, 2022, 2023, 2025 papers).

Practice MCQs

Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.

MCQ 1Easy RecallPractice

Phenol reacts with bromine water at room temperature to form:

MCQ 2Easy RecallPractice

Which catalyst is required for bromination of phenol with bromine water?

MCQ 3Easy RecallPractice

The product of Kolbe's reaction (sodium phenoxide + CO₂ at 125°C under pressure, followed by acidification) is:

MCQ 4Direct ApplicationPractice

To obtain predominantly para-bromophenol from phenol, the appropriate reagent and solvent are:

MCQ 5Direct ApplicationPractice

Phenol is treated with CHCl₃ and NaOH, followed by acidification. The product is:

MCQ 6Direct ApplicationPractice

Among the following substituted phenols, which is the most acidic?

MCQ 7Concept TrapPractice

Phenol undergoes electrophilic substitution much more readily than benzene because:

MCQ 8CalculationPractice

Phenol is first treated with dilute HNO₃ at low temperature, and the para-nitrophenol product is then isolated. If this para-nitrophenol is now brominated with Br₂/H₂O, the expected product is:

Worked Example

  1. 1

    Given

    Four substituted phenols: phenol (C₆H₅OH), p-nitrophenol (O₂N–C₆H₄–OH), p-cresol (CH₃–C₆H₄–OH), p-chlorophenol (Cl–C₆H₄–OH).

  2. 2

    Required

    Decreasing order of acidity.

  3. 3

    Concept

    Acidity of phenols depends on stability of the phenoxide ion formed after proton loss. Electron-withdrawing groups (EWG) stabilise phenoxide by dispersing negative charge → increase acidity. Electron-donating groups (EDG) destabilise phenoxide → decrease acidity. Both resonance and inductive effects must be considered; ignoring one leads to wrong ordering.

  4. 4

    Principle applied

    For para-substituents on phenol: - –NO₂ is strongly electron-withdrawing by both –I and –R effects → strongly stabilises phenoxide. - –Cl is electron-withdrawing by –I effect but weakly electron-donating by +R effect. Net effect at the para position: weak electron withdrawal → mildly stabilises phenoxide. - –CH₃ is electron-donating by hyperconjugation and +I effect → destabilises phenoxide.

  5. 5

    Analysis (substituent by substituent)

    - p-Nitrophenol: –NO₂ has strong –I and –R → most stable phenoxide → most acidic. - p-Chlorophenol: –Cl has –I (moderate) and weak +R → net mild stabilisation → more acidic than phenol. - Phenol: no substituent → reference point. - p-Cresol: –CH₃ has +I and hyperconjugation → destabilises phenoxide → least acidic.

  6. 6

    Ordering

    p-Nitrophenol > p-Chlorophenol > Phenol > p-Cresol

  7. 7

    Final answer

    Decreasing acidity: **p-Nitrophenol > p-Chlorophenol > Phenol > p-Cresol**

  8. 8

    Common trap

    The documented distractor pattern here is "ignores resonance-vs-induction balance." A student relying only on electronegativity might rank –Cl as strongly acidifying (comparable to –NO₂) or might forget that –CH₃ is electron-donating. The resonance effect of –NO₂ withdrawing through the ring is far stronger than the inductive effect of –Cl alone.

  9. 9

    Similar NEET-style question

    Arrange in increasing order of acidity: 2,4-dinitrophenol, m-cresol, phenol, p-fluorophenol. (Same principle — EWG/EDG effects on phenoxide stability — but with a disubstituted phenol and a halogen at a different position.) ---

Before solving, remember these

Key Fact

Oxides of nitrogen — structures and oxidation states

N forms eight oxides: N₂O (+1), NO (+2), N₂O₃ (+3), NO₂/N₂O₄ (+4), N₂O₅ (+5), plus N₂O₂²⁻, NO₃⁻. NO is a brown paramagnetic gas with odd electron. NO₂ is also paramagnetic (odd electron, bent). N₂O₄ is diamagnetic dimer of NO₂. N₂O₅ is solid acid anhydride of HNO₃. Key acid anhydrides: N₂O₃ of HNO₂; N₂O₅ of HNO₃.

-- NCERT Class 12 Chemistry, Ch. 7, p. 8

Formulas

pKa of carboxylic acid

Stronger acid than phenol due to more effective resonance over two equivalent oxygens. EWG substituents (Cl, NO2) increase acidity.

SymbolQuantitySI Unit
pKa-log Ka-

Valid when

  • Aqueous solution
  • α-substituent effects strongest

pKa of phenol vs aliphatic alcohol

Phenols ~10⁶× more acidic than aliphatic alcohols due to resonance stabilisation of phenoxide ion. Electron-withdrawing substituents lower pKa further.

SymbolQuantitySI Unit
pKa-log Ka-

Valid when

  • Aqueous solution
  • Substituent effects shift values

Exam Traps & Common Mistakes

These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.

Trap

Oxidation product depends on reagent strength

Category: Organic Reaction Conditions

1° alcohol: PCC/PDC → aldehyde (stops). KMnO4/K2Cr2O7 in acidic → carboxylic acid (continues). 2° alcohol: any oxidiser → ketone. 3° alcohol: not oxidised by ordinary reagents.

When it triggers

Question gives 1° alcohol oxidation with specified reagent.

How to avoid

PCC, PDC, Swern, DMP: mild → stop at aldehyde. KMnO4, K2Cr2O7, CrO3, jones: strong → carboxylic acid. Reagent choice matches desired product.

Past Year Questions

20 questions from NEET 2021, 2022, 2023, 2024, 2025. Answers verified against NTA official keys.

NEET 2024Revised key

Given below are two statements: Statement I : Aniline does not undergo Friedel-Crafts alkylation reaction. Statement II : Aniline cannot be prepared through Gabriel synthesis . In the light of the above statements, choose the correct answer from the options given below:

1Both statement I and Statement II are true
2Both Statement I and Statement II are false
3Statement I is correct but Statement II is false
4Statement I is incorrect but Statement II is true
NTA Answer: Option 1(revised_final)
NEET 2024Revised key

Fehling’s solution ‘A’ is

1aqueous copper sulphate
2alkaline copper sulphate
3alkaline solution of sodium potassium tartrate (Rochelle’s salt)
4aqueous sodium citrate
NTA Answer: Option 1(revised_final)
NEET 2022

Given below are two statements: Statement I: In Lucas test, primary, secondary and tertiary alcohols are distinguished on the basis of their reactivity with conc. HCl + ZnCl , known as Lucas Reagent. 2 Statement II: Primary alcohols are most reactive and immediately produce turbidity at room temperature on reaction with Lucas Reagent. In the light of the above statements, choose the most appropriate answer from the options given below:

1Statement I is incorrect but Statement II is correct
2Both Statement I and Statement II are correct
3Both Statement I and Statement II are incorrect
4Statement I is correct but Statement II is incorrect
NTA Answer: Option 4(final)
NEET 2021

Match List-I with List-II List-I List-II CO, HCI (a) (i) Hell-Volhard-Zelinsky Anhyd. AlCl/ Cu3Cl reaction O R — C — CH + (b) (ii) Gattermann-Koch 3 NaOX reaction R — CH — OH (c) (iii) Haloform reaction 2 +R′ COOH Conc. HSO 2 4 (d) R — CH COOH (iv) Esterification 2 ( II) X /Red P 2 (ii) HO 2 Choose the correct answer from the options given below.

1(a) - (ii), (b) - (iii), (c) - (iv), (d) - (i)
2(a) - (iv), (b) - (i), (c) - (ii), (d) - (iii)
3(a) - (iii), (b) - (ii), (c) - (i), (d) - (iv)
4(a) - (i), (b) - (iv), (c) - (iii), (d) - (ii)
NTA Answer: Option 1(final)

How NEET usually asks this

Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.

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